TITLE: On $\int_0^{\pi} \operatorname{Si}^n(x) \ \mathrm{d}x $
QUESTION [17 upvotes]: The Sine Integral, $\operatorname{Si}(x)$, is defined as $$\operatorname{Si}(x)=\int_0^x \frac{\sin t}{t} \ \mathrm{d}t $$
It follows that $$\int_0^{\pi} \operatorname{Si}(x) \ \mathrm{d}x = \bigg[x \operatorname{Si}(x) \bigg]_0^{\pi}-\int_0^{\pi} \sin x \ \mathrm{d}x = \pi \operatorname{Si} (\pi)-2$$ and further $$\int_0^{\pi} \operatorname{Si}^2 (x) \ \mathrm{d}x = \bigg[ x \operatorname{Si}^2(x) \bigg]_0^{\pi} -2\int_0^{\pi} \operatorname{Si} (x) \sin x \ \mathrm{d}x \\ = \pi \operatorname{Si}^2(\pi) -2 \left( \big[ -\cos x \operatorname{Si} (x) \big]_0^{\pi}+\int_0^{\pi} \frac{\sin x \cos x}{x} \ \mathrm{d}x \right)\\ = \pi \operatorname{Si}^2(\pi)-2 \operatorname{Si} (\pi)-\int_{2x=0}^{2\pi}\frac{\sin 2x}{2x} \ \mathrm{d}(2x)\\ = \pi\operatorname{Si}^2(\pi)-2 \operatorname{Si} (\pi)-\operatorname{Si} (2\pi)$$
Following a similar procedure on $\int_0^{\pi} \operatorname{Si}^3(x) \ \mathrm{d}x$ doesn’t quite work, because it requires evaluating the integral $\int_0^{\pi} \operatorname{Si}^2(x) \sin x \ \mathrm{d}x$, or equivalently, $\int_0^{\pi} \frac{\operatorname{Si}(x) \sin 2x}{x} \ \mathrm{d}x$.
This leads me to ask, can (and how can) $\int_0^{\pi} \operatorname{Si}^3(x) \ \mathrm{d}x$ be expressed in terms of elementary or special functions? How about the general case, i.e.  $$\color{purple}{\int_0^{\pi} \operatorname{Si}^n(x) \ \mathrm{d}x} \; ? $$

REPLY [2 votes]: Here is an alternative series expression for the integral
$$i = \int_{0}^{\pi} \frac{\sin(2x)}{x} \text{Si}(x )\,dx\tag{1}$$
which I obtained from replacing the $\sin$ by its power series and integrating termwise. A similar series was provided by TheSimpliFire.
From
$$c(n) = \int_0^{\pi } x^{2 n} \text{Si}(x) \, dx \\
= \frac{\pi ^{2 n+1} \left(2 (n+1) \text{Si}(\pi )-\pi  \, _1F_2\left(n+1;\frac{3}{2},n+2;-\frac{\pi ^2}{4}\right)\right)}{2 (n+1) (2 n+1)}\tag{2}$$
which I have derived by looking for a recursive solution for $c(n)$ (or alternatively simply emloying Mathematica), I get
$$i = s := s_1 + s_2$$
where
$$s_1 =\sum _{n=0}^{\infty } \frac{(-1)^n (2 \pi )^{2 n+1} \text{Si}(\pi )}{(2 n+1) (2 n+1)!} =\text{Si}(\pi ) \text{Si}(2 \pi ) \tag{3}$$
and
$$s_2 =-\pi\sum _{n=0}^{\infty } \frac{(-1)^n (2 \pi )^{2 n+1} \, _1F_2\left(n+1;\frac{3}{2},n+2;-\frac{\pi ^2}{4}\right)}{2 (n+1) (2 n+1) (2 n+1)!}\tag{4}$$
The hypergeometric function $s_2$ is related to Bessel functions as can be seen from the first four summands
$$\begin{align}\left\{-4,\frac{1}{9} (-2) \left(4+\sqrt{2} \pi ^3 J_{-\frac{1}{2}}(\pi )+2 \sqrt{2} \pi ^2 J_{-\frac{3}{2}}(\pi )\right),\frac{1}{75} (-2) \left(48-\sqrt{2} \pi ^4 J_{-\frac{3}{2}}(\pi )+\sqrt{2} \pi ^5 J_{-\frac{5}{2}}(\pi )-8 \sqrt{2} \pi ^3 J_{-\frac{5}{2}}(\pi )\right),\frac{4 \left(-1440+\sqrt{2} \pi ^5 \left(\pi ^2-9\right) J_{-\frac{5}{2}}(\pi )+3 \sqrt{2} \pi ^4 \left(\pi ^2-16\right) J_{-\frac{7}{2}}(\pi )\right)}{2205}\right\}\end{align}\tag{4a}$$
They simplify, however, to these curious polynomials in $\pi$:
$$\begin{align}\left\{-4,\frac{4 \pi ^2}{9}-\frac{16}{9},-\frac{64}{25}+\frac{16 \pi ^2}{25}-\frac{4 \pi ^4}{75},\\
-\frac{256}{49}+\frac{64 \pi ^2}{49}-\frac{16 \pi ^4}{147}+\frac{8 \pi ^6}{2205}\right\}\end{align}\tag{4b}$$
(As expected) I was not able to calculate $s_2$ analytically.
But notice that the hypergeometric function appearing here is of smaller degree than that in the sum in 1 so that maybe some progress is possible.
EDIT
The sum can also be written with a hypergeometric function with smaller parameters as
$$\begin{align}s=\left(\text{Si}(\pi )-\frac{1}{\pi }\right) \text{Si}(2 \pi )-\sum _{n=0}^{\infty } \frac{(-1)^n \left(2 (2 \pi )^{2 n}\right) \, _1F_2\left(n;\frac{1}{2},n+1;-\frac{\pi ^2}{4}\right)}{(2 n+1) (2 n+1)!}\end{align}\tag{5}$$
$\, _1F_2$ admits the representations
$$\begin{align}& \, _1F_2\left(n;\frac{1}{2},n+1;-\frac{\pi ^2}{4}\right)\\
& = n \sqrt{\pi }  \sum _{k=0}^{\infty } \frac{1}{ (k+n) \Gamma \left(k+\frac{1}{2}\right)}\frac{z^k}{k!}|_{z\to -(\frac{\pi}{2})^2}\\
& =n \int_{0}^{1} x^{n-1}\cos(\pi \sqrt{x})\,dx=2 n \int_0^1  y^{2 n-1} \cos (\pi  y) \, dy\end{align}\tag{6}$$